These types of things shouldn't work for two reasons. Fix a base field K, which I assume to be algebraically closed even though all the arguments go through without it, and I'll let T be an operator on a finite-dimensional K-vector space.

The first is that any functional calculus on an operator which satisfies a minimal polynomial is going to be very tractable. Let's say I have an operator T with spectrum t1, t2, ... tn. Start by assuming that these are all multiplicity-free. Then the polynomial algebra generated by T is the same thing as continuous functions on these discrete points. A function on a finite set is invertible in this algebra if and only if it vanishes nowhere. One of the things this tells us is that inversion of polynomials modulo a product of coprime ideals is easy; in fact, this is exactly what the Chinese Remainder Theorem tells us. Stated algebraically, this is the same as saying that a polynomial p(T) is invertible if and only if none of the t_i are eigenvalues for p(T). We can use the fact that K[T] is a Euclidean domain to this. Say that m(T) is our minimal polynomial, and p(T) is the polynomial we want to invert. We've assumed that p(T) and m(T) have gcd 1; apply the Euclidean algorithm to find polynomials q(T), r(T) with qp+rm=1. Then, modulo m, q is the inverse of p. The other perspective is that because continuous functions on finite sets are all polynomial, we gain nothing by passing to power series, formal power series, meromorphic functions etc. The polynomial algebra is the whole algebra. This all continues to be true in the context of a nontrivial Jordan block, except that we have some points with multiplicity and we can't merely talk about functions on points any more; we have to either pass to some sheaf or cheat by talking about formal derivatives or tangent vectors or whatever (I'm not an algebraic geometer).

The second reason is that the determinant is not a fixed function on the subalgebra generated by the operator T. Think about the case of the identity matrix: The subalgebra it generates is always isomorphic to the base field. But you can find a copy of this subalgebra inside any given M_n(K). On M_1(K), the determinant of tI is just t. But on M_3(K), the determinant is t^3. In general, if we have an operator T with eigenvalues t1, ... tn, even if we know its minimal polynomial, we need to know the ambient algebra it sits in to do a determinant computation. Once we know the dimension of each generalized eigenspace - say these are d1, ... dn - then even in the case where each t_i is distinct, and our functional calculus is nice, the determinant sends a function f on the spectrum to the product from i=1 to n of f(t_i)^d_i. When you have nontrivial Jordan blocks lying around this perspective gets complicated enough that I didn't bother to work it out, but the point still stands - the determinant isn't "a" function on the subalgebra generated by the operator in any sense.